'light of wavelength 2000 Å falls on a metal surface.the work function is 4.2eV.

what is the kinetic energy of the fastest and slowest emmited photoelectrons?

KE = hf - Φ

We know hf and Φ, so we can just plug and chug, right?

v = λf
f = v/λ = c/λ = (3 × 10⁸ m/s) / (2 × 10⁻⁷ m) = 15 Hz

KE = hf - Φ
KE = (6.626068 × 10⁻³⁴ m² kg / s)(15 Hz) - ( 6.72914114 × 10⁻¹⁹ J)
KE = -6.72914 × 10⁻¹⁹ J ≈ 4.2 eV

KE = (1/2) mv²

6.72914 × 10⁻¹⁹ J = (1/2) (9.10938188 × 10⁻³¹ kg) v²
v ≈ 1.21549 × 10⁶ m/s = 0.004c

Oh sorry I didn't see the min and max part. Yes.

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λ =2000 Å = 2e-7 m
E = hc/ λ = 6.626e-34 * 3e8/2e-7 =9.939e-19 J
V² = (E – W) / (2m) = {9.939e-19 – (4.2*1.602e-19)} / (2* 9.1e-31)
V max = 5.94 e+5 m/s
V minimum is zero.

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